I think that the problem point for set A is 0. I believe that f(x) is continuous there because I drew an epsilon in my mind around f(x) on the y-axis and imagined that I could find a delta such that not only the (x not in A) values would map into it (obviously) but that I could choose delta such that the 1/n values were far enough down, and thus in the epsilon N. I hope I am right on this. The problem points are hard to figure out.

With B I am using the same reasoning. This time though we have discontinuous at x=0 because the f(x) values are at value 1 while there are always values in your delta N such that f(x)=0. But at x=1, f(x) is getting as close as we want it to get to f(x)=0.